1. ## simple one

let p be odd prime p=8k+r 1<=r<=7
prove that

$(-1)^{(p^2 - 1)/8} = (-1)^{(r^2 - 1)/8}$

2. Originally Posted by yeke
let p be odd prime p=8k+r 1<=r<=7
prove that

(-1)^((p^2-1)/8) = (-1)((r^2-1)/8)
Are you sure this isn't
$(-1)^{(p^2 - 1)/8} = (-1)^{(r^2 - 1)/8}$

Form p^2 - 1 = 64k^2 + 16kr + (r^2 - 1). Divide this by 8 and plug it into the LHS of your original statement and see what cancels.

-Dan

3. Originally Posted by topsquark
Are you sure this isn't
$(-1)^{(p^2 - 1)/8} = (-1)^{(r^2 - 1)/8}$

Form p^2 - 1 = 64k^2 + 16kr + (r^2 - 1). Divide this by 8 and plug it into the LHS of your original statement and see what cancels.

-Dan
yes thats it. but i didnt understand about forming p^2 - 1... how am i gonna do that?

4. Originally Posted by yeke
yes thats it. but i didnt understand about forming p^2 - 1... how am i gonna do that?
$p = 8k + r$

$p^2 - 1 = (8k + r)^2 - 1 = 64k^2 + 16kr + r^2 - 1$

So

$(-1)^{(p^2 - 1)/8} = (-1)^{(64k^2 + 16kr + r^2 - 1)/8}$

$= (-1)^{8k^2 + 2kr + (r^2 - 1)/8}$

$= (-1)^{8k^2}(-1)^{2kr}(-1)^{(r^2 - 1)/8}$

$= (1)(1)(-1)^{(r^2 - 1)/8} = (-1)^{(r^2 - 1)/8}$

-Dan